The grades on a geometry midterm at Almond are normally distributed with $\mu = 83$ and $\sigma = 3.0$. Umaima earned a n $85$ on the exam. Find the z-score for Umaima's exam grade. Round to two decimal places.
A z-score is defined as the number of standard deviations a specific point is away from the mean We can calculate the z-score for Umaima's exam grade by subtracting the mean $(\mu)$ from her grade and then dividing by the standard deviation $(\sigma)$ $ { z = \dfrac{x - {\mu}}{{\sigma}}} $ $ { z = \dfrac{85 - {83}}{{3.0}}} $ ${ z \approx 0.67}$ The z-score is $0.67$. In other words, Umaima's score was $0.67$ standard deviations above the mean.